Question: Let $f$ be a polynomial function and let $f'$, its derivative, be defined as $f'(x)=-x(x+2)(x-2)$. At how many points does the graph of $f$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Explanation: We can find the relative extrema (i.e. minima and maxima) of $f$ by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $f'(x)=-x(x+2)(x-2)$. $f'(x)=0$ for $x=-2,0,2$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-2$, $x=0$, and $x=2$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $(-\infty, \llap{-}2)$ $(\ \ \llap{-}2,0)$ $(0,2)$ $(2,\infty)$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $(-\infty,-2)$ $x=-3$ $f'(-3)=15>0$ $f$ is increasing $\nearrow$ $(-2,0)$ $x=-1$ $f'(-1)=-3<0$ $f$ is decreasing $\searrow$ $(0,2)$ $x=1$ $f'(1)=3>0$ $f$ is increasing $\nearrow$ $(2,\infty)$ $x=3$ $f'(3)=-15<0$ $f$ is decreasing $\searrow$ Now let's look at the critical points: $x$ Before After Verdict $-2$ $\nearrow$ $\searrow$ Maximum $0$ $\searrow$ $\nearrow$ Minimum $2$ $\nearrow$ $\searrow$ Maximum Now we can see that $f$ has two relative maxima.